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17 Mar 2020, 04:51
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B
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Difficulty:
95%
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Question Stats:
53% (02:58) correct
47%
(02:51)
wrong
based on 194
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If n is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is a positive integer, what is the value of n?
(1) n is a factor of a prime number
(2) n < 3
Are You Up For the Challenge: 700 Level Questions
M36-53
(1) n is a factor of a prime number
(2) n < 3
Are You Up For the Challenge: 700 Level Questions
M36-53
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08 Nov 2021, 05:47
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Bunuel
Official Solution:
If \(n\) is a positive integer, and \(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\) is an integer, what is the value of \(n\)?
Simplify given expression:
\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}=\)
\(=\sqrt{2*3^2*5*7^2*7^{(n-1)} - 2*3^4*5*7^{(n - 1)}}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(7^2 - 3^2)}=\)
\(=\sqrt{2*3^2*5*7^{(n-1)}(2^3*5)}=\)
\(=\sqrt{2^4*3^2*5^2*7^{(n-1)}}=\)
\(=2^2*3*5*\sqrt{7^{(n-1)}}\)
So, we are given that \(2^2*3*5\sqrt{7^{(n-1)}}\) is an integer.
Notice that since \(n\) is a positive integer, then \(7^{(n-1)}\) is also a positive integer.
Now, the square root of any positive integer is either an integer or an irrational number. Meaning that, \(\sqrt{positive \ integer}\) cannot be a fraction, for example it cannot equal to \(\frac{1}{2}, \ \frac{3}{7}, \ \frac{19}{2}, \ \frac{1}{60}\) ... It MUST be an integer (1, 2, 3, ...) or an irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{7}\), ...). Thus, since \(7^{(n-1)}\) is a positive integer then \(\sqrt{7^{(n-1)}}\) is either an integer itself or an irrational number.
Therefore, for \(2^2*3*5*\sqrt{7^{(n-1)}}\) to be an integer, \(\sqrt{7^{(n-1)}}\) must be an integer. \(\sqrt{7^{(n-1)}}\) is an integer for (positive) ODD values of \(n\).
So, from above, \(n\) is a positive odd integer: 1, 3, 5, 7, ...
(1) \(n\) is a factor of a prime number
\(n\) can be 1 or any odd prime. Not sufficient.
(2) \(n < 3\)
There is only one positive odd number less than 3, namely 1. So, \(n = 1\). Sufficient.
Answer: B
General Discussion
Updated on: 17 Mar 2020, 07:04
Originally posted by CareerGeek on 17 Mar 2020, 05:08.
Last edited by CareerGeek on 17 Mar 2020, 07:04, edited 1 time in total.
Last edited by CareerGeek on 17 Mar 2020, 07:04, edited 1 time in total.
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Bunuel
\(\sqrt{45*14*7^n - 15*7^{(n - 1)}*54}\)
= \(\sqrt{2*3^2*5*7^{n + 1} - 2*3^4*5*7^{(n - 1)}}\)
= \(\sqrt{2*3^2*5*7^2*7^{n - 1} - 2*3^4*5*7^{(n - 1)}}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(7^2 - 3^2)}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(40)}\)
= \(\sqrt{2*3^2*5*7^{n - 1}(2^3*5)}\)
= \(\sqrt{2^4*3^2*5^2*7^{n - 1}}\)
= \(2^2*3*5\sqrt{7^{n - 1}}\)
= \(60\sqrt{7^{n - 1}}\) is a positive integer
The above is a positive integer for all odd values of n = {1, 3, 5, 7,, 9, . . . . . . }
(1) \(n\) is a factor of a prime number
--> Possible values of \(n = {1, 2, 3, 5, 7, 9, . . . . . .}\) --> Not always Odd --> Insufficient
(2) \(n < 3\)
--> Possible values of \(n = 1\) --> Sufficient
Option B
